# Open mapping theorem (functional analysis)

Suppose

${displaystyle A:Xto Y}$

is a surjective continuous linear operator. In order to prove that

${displaystyle A}$

is an open map, it is sufficient to show that

${displaystyle A}$

maps the open unit ball in

${displaystyle X}$

to a neighborhood of the origin of

${displaystyle Y.}$

Let

${displaystyle U=B_{1}^{X}(0),V=B_{1}^{Y}(0).}$

Then

${displaystyle X=bigcup _{kin mathbb {N} }kU.}$

Since

${displaystyle A}$

is surjective:

${displaystyle Y=A(X)=Aleft(bigcup _{kin mathbb {N} }kUright)=bigcup _{kin mathbb {N} }A(kU).}$

But

${displaystyle Y}$

is Banach so by Baire’s category theorem

${displaystyle exists kin mathbb {N} :qquad left({overline {A(kU)}}right)^{circ }neq varnothing .}$

That is, we have

${displaystyle cin Y}$

and

${displaystyle r>0}$

${displaystyle B_{r}(c)subseteq left({overline {A(kU)}}right)^{circ }subseteq {overline {A(kU)}}.}$

Let

${displaystyle vin V,}$

then

${displaystyle c,c+rvin B_{r}(c)subseteq {overline {A(kU)}}.}$

By continuity of addition and linearity, the difference

${displaystyle rv}$

satisfies

${displaystyle rvin {overline {A(kU)}}+{overline {A(kU)}}subseteq {overline {A(kU)+A(kU)}}subseteq {overline {A(2kU)}},}$

and by linearity again,

${displaystyle Vsubseteq {overline {A(LU)}}}$

where we have set

${displaystyle L=2k/r.}$

It follows that for all

${displaystyle yin Y}$

and all

${displaystyle epsilon >0,}$

${displaystyle xin X}$

such that

${displaystyle qquad |x|_{X}leq L|y|_{Y}quad {text{and}}quad |y-Ax|_{Y}$

Our next goal is to show that

${displaystyle Vsubseteq A(2LU).}$

Let

${displaystyle yin V.}$

By (1), there is some

${displaystyle x_{1}}$

with

${displaystyle left|x_{1}right|$

and

${displaystyle left|y-Ax_{1}right|<1/2.}$

Define a sequence

${displaystyle left(x_{n}right)}$

inductively as follows.
Assume:

${displaystyle |x_{n}|<{frac {L}{2^{n-1}}}quad {text{and}}quad left|y-Aleft(x_{1}+x_{2}+cdots +x_{n}right)right|<{frac {1}{2^{n}}}.qquad (2)}$

Then by (1) we can pick

${displaystyle x_{n+1}}$

so that:

${displaystyle |x_{n+1}|<{frac {L}{2^{n}}}quad {text{and}}quad left|y-Aleft(x_{1}+x_{2}+cdots +x_{n}right)-Aleft(x_{n+1}right)right|<{frac {1}{2^{n+1}}},}$

so (2) is satisfied for

${displaystyle x_{n+1}.}$

Let

${displaystyle s_{n}=x_{1}+x_{2}+cdots +x_{n}.}$

From the first inequality in (2),

${displaystyle left(s_{n}right)}$

is a Cauchy sequence, and since

${displaystyle X}$

is complete,

${displaystyle s_{n}}$

converges to some

${displaystyle xin X.}$

By (2), the sequence

${displaystyle As_{n}}$

tends to

${displaystyle y}$

and so

${displaystyle Ax=y}$

by continuity of

${displaystyle A.}$

Also,

${displaystyle |x|=lim _{nto infty }|s_{n}|leq sum _{n=1}^{infty }|x_{n}|<2L.}$

This shows that

${displaystyle y}$

belongs to

${displaystyle A(2LU),}$

so

${displaystyle Vsubseteq A(2LU)}$

as claimed.
Thus the image

${displaystyle A(U)}$

of the unit ball in

${displaystyle X}$

contains the open ball

${displaystyle V/2L}$

of

${displaystyle Y.}$

Hence,

${displaystyle A(U)}$

is a neighborhood of the origin in

${displaystyle Y,}$

and this concludes the proof.